In the Auger process,an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive,calculate the kinetic energy of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from an $n = 2$ to $n = 1$ transition.

(D) The energy of an electron in a hydrogen-like atom is given by $E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \ eV$.
For Chromium $(Z = 24)$:
Energy of $n = 2$ state: $E_{2} = -13.6 \times \frac{24^{2}}{2^{2}} = -13.6 \times 144 = -1958.4 \ eV$.
Energy of $n = 1$ state: $E_{1} = -13.6 \times \frac{24^{2}}{1^{2}} = -13.6 \times 576 = -7833.6 \ eV$.
The energy released during the $n = 2$ to $n = 1$ transition is $\Delta E = E_{2} - E_{1} = -1958.4 - (-7833.6) = 5875.2 \ eV$.
This energy is transferred to an electron in the $n = 4$ state.
The energy of the $n = 4$ state is $E_{4} = -13.6 \times \frac{24^{2}}{4^{2}} = -13.6 \times 36 = -489.6 \ eV$.
The kinetic energy $K$ of the ejected Auger electron is $K = \Delta E - |E_{4}| = 5875.2 - 489.6 = 5385.6 \ eV$.